What determines the time of one half-cycle of the oscillation? Does this depend on VCC?
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The time of the half-cycle described previously is the time
that it takes the right end of C1 to charge from –(VCC – (VBE
+ VCE(SAT))) to +VBE.
Now, keep in mind that the capacitor is charging “toward” +VCC, but it gets halted by the B-E junction of Q2 at +VBE. This charging is occuring at a rate determined by the time constant C1 × R2, and we’re basically interested in the time that it takes to move halfway from its starting value to its final value. This works out to –ln(0.5), or 0.693 times the R-C time constant. As long as VCC >> VBE, the time does not depend on VCC. That isn’t to say, however, that VCC can be arbitrarily large. If it exceeds the reverse-breakdown voltage of the transistors’ B-E junctions, current will flow and perturb the timing.
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As long as VCC >> VBE, the time does not depend on VCC. That isn’t to say, however, that VCC can be arbitrarily large. If it exceeds the reverse-breakdown voltage of the transistors’ B-E junctions, current will flow and perturb the timing.
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